Problem: Solve for $x$ and $y$ by deriving an expression for $y$ from the second equation, and substituting it back into the first equation. $\begin{align*}6x+8y &= 4 \\ 4x-8y &= 6\end{align*}$
Explanation: Begin by moving the $x$ -term in the second equation to the right side of the equation. $-8y = -4x+6$ Divide both sides by $-8$ to isolate $y$ $y = {\dfrac{1}{2}x - \dfrac{3}{4}}$ Substitute this expression for $y$ in the first equation. $6x+8({\dfrac{1}{2}x - \dfrac{3}{4}}) = 4$ $6x + 4x - 6 = 4$ Simplify by combining terms, then solve for $x$ $10x - 6 = 4$ $10x = 10$ $x = 1$ Substitute $1$ for $x$ back into the top equation. $6( 1)+8y = 4$ $6+8y = 4$ $8y = -2$ $y = -\dfrac{1}{4}$ The solution is $\enspace x = 1, \enspace y = -\dfrac{1}{4}$.